Question: Divide the following complex numbers. $ \dfrac{-12+8i}{2-2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+2i}$ $ \dfrac{-12+8i}{2-2i} = \dfrac{-12+8i}{2-2i} \cdot \dfrac{{2+2i}}{{2+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-12+8i) \cdot (2+2i)} {(2-2i) \cdot (2+2i)} = \dfrac{(-12+8i) \cdot (2+2i)} {2^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-12+8i) \cdot (2+2i)} {(2)^2 - (-2i)^2} = $ $ \dfrac{(-12+8i) \cdot (2+2i)} {4 + 4} = $ $ \dfrac{(-12+8i) \cdot (2+2i)} {8} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-12+8i}) \cdot ({2+2i})} {8} = $ $ \dfrac{{-12} \cdot {2} + {8} \cdot {2 i} + {-12} \cdot {2 i} + {8} \cdot {2 i^2}} {8} $ Evaluate each product of two numbers. $ \dfrac{-24 + 16i - 24i + 16 i^2} {8} $ Finally, simplify the fraction. $ \dfrac{-24 + 16i - 24i - 16} {8} = \dfrac{-40 - 8i} {8} = -5-i $